|
%I #50 Apr 21 2022 13:13:24
%S 2,4,6,6,8,8,8,12,12,8,12,12,8,16,20,10,12,12,12,24,16,8,16,24,12,16,
%T 24,12,16,16,12,24,16,16,36,18,8,16,32,16,16,16,12,36,24,8,20,30,18,
%U 24,24,12,16,32,32,32,16,8,24,24,8,24,42,28,32,16,12,24,32,16,24,24,8,24,36
%N Product of tau values for consecutive integers.
%C Number of divisors of the n-th oblong number. - _Ray Chandler_, Jun 23 2008
%C Number of positive solutions (x,y) for which n/x + (n+1)/y = 1. - _Michel Lagneau_, Jan 16 2014
%C Number of positive solutions for which 1/p + 1/q + 1/(p*q) = 1/n; set p=x and q=y-1 in the solutions (x,y) in the comment above. - _Mo Li_, Apr 27 2021
%C a(n) is the maximum number of b > 0, which allows us to write (n+1)^2 as a sum of n+1 parts. Each part is of the form b^c and c is an integer >= 0 independent for each part. For n = 2 this is 3^2 = 2^2 + 2^2 + 2^0 = 3^1 + 3^1 + 3^1 = 4^1 + 4^1 + 4^0 = 7^1 + 7^0 + 7^0, b = 2;3;4;7 and a(2) = 4. It is conjectured that for all n the number of possible b reaches a(n). - _Thomas Scheuerle_, Jan 12 2022
%H Ray Chandler, <a href="/A092517/b092517.txt">Table of n, a(n) for n = 1..10000</a>
%F a(n) = A000005(n)*A000005(n+1) = A000005(n*(n+1)) = A000005(A002378(n)) = 2*A063123(n).
%p with(numtheory): seq(tau(n)*tau(n+1),n=1..73); # _Zerinvary Lajos_, Jan 22 2007
%t Table[DivisorSigma[0,n^2+n],{n,100}] (* _Giorgos Kalogeropoulos_, Apr 28 2021 *)
%t Times@@#&/@Partition[DivisorSigma[0,Range[80]],2,1] (* _Harvey P. Dale_, Apr 21 2022 *)
%o (Magma) [ NumberOfDivisors(n^2+n) : n in [1..100]]; // _Vincenzo Librandi_, Apr 03 2011
%o (PARI) a(n) = numdiv(n^2+n); \\ _Michel Marcus_, Jan 11 2020
%o (Python)
%o from sympy import divisor_count
%o def A092517(n): return divisor_count(n)*divisor_count(n+1) # _Chai Wah Wu_, Jan 06 2022
%Y Cf. A000005, A002378, A063123, A063440, A083539, A123000.
%K nonn
%O 1,1
%A _Reinhard Zumkeller_, Apr 06 2004
%E Extended by _Ray Chandler_, Jun 23 2008
|
